Cantor Confusion
in [Math]
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From: mueckenh on 23 Mar 2007 09:15 On 22 Mrz., 16:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174563273.137227.294...(a)y66g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 21 Mrz., 16:16, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1174481667.234070.21...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > > Yes. But the path-lengths are natural numbers, and there is no > > > > > > infinite natural number. > > > > > > > > > > The path-length of a non-terminating path is *not* a natural number. > > > > > > > > How then can it be contained in a set which contains only sets of > > > > finite paths? > > > > > > It is *not* an element of a set which contains only sets of finite paths. > > > Pray get your statements correct. A set that contains only sets of finite > > > paths has as elements sets of finite paths. Not paths. > > > > We can form the union of the sets. Then we get a set of finite paths. > > Yes, we can do that. And then we get a set of finite paths that does *not* > contain a non-terminating path as an element, and also not the path-length > of such a path as an element. So what now? This set of paths forms a tree which does not differ by any node from the complete tree T(oo). > Again "contain". But, so what? I fail to see the relevance to the > path-length of non-terminating paths. To spell it out clearly: If you take the infinite union U(T(n)) of all finite trees T(n), then you do not have any infinite path. > > > > > why then do you believe that an infinite path (number) is contained in > > > > the union of sets of finite paths (numbers). > > > > > > It is not an element of the union of sets of finite paths. Why do you > > > think that I do think so? The union of sets of finite paths is a set > > > of finite paths, so there is no infinite path in it as element. > > > > Therefore, after having the union of all paths which are elements of > > finite trees, there is no infinite path. > > Wrong. In the first place, paths are *not* elements of trees according to > your definitions. Paths are *subsets* of trees according to them. Pray > use the correct wording. OK. Subsets. > Second, the union of all paths etc., is *not* a > path, it is just an infinite set of nodes, the same as the union of all > trees. The paths are in it as subsets, as are the infinite paths. The union of all *finite* paths gives the same infinite set of nodes as the union of all finite trees. But there is no infinite path involved / required. > > > > > > Where is the infinite element? > > > > > > > > You said: "The path-length of a non-terminating path is *not* a > > > > natural number." And you find it in a set of sets of finite paths. > > > > > > Eh? N is the union of the sets of initial segments, the cardinal number > > > of N is not a natural number. Where is the difference? The path-length > > > is the cardinal number of the set that represents a path (actually one > > > less if the path is finite). > > > > No. You are in error. The cardinal number of an infinite set of finite > > paths is infinite. But that does not produce any path of infinite > > length. > > But I do not state that it does. The infinite paths are *unions* of finite > paths, where each finite path is an initial segment of the infinite path. > I am not talking about the cardinality of a set of path, but of the > cardinality of a path (the number of nodes in it). That does not mean > that an infinite set of paths contains an infinite path, that would be > false indeed. The union of all finite paths is an infinite union of finite paths, but not an infinite path. The set of all negative unit fractions is infinite but neither any member nor the whole set does cover zero. > > > > > > > (3) There is no node in the tree (including every node of the paths > > > > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > > > > many paths arrive at or start off from or cross through. > > > > > > > > > > No. From the root node of the tree there start uncountably many paths. > > > > > > > > But they do never separate into uncountably many paths. > > > > > > At each node, uncountably many paths go to the left and uncountably many > > > paths go to the right. > > > > And everything which belongs to the mathematics of real numbers of [0, > > 1] happens within the tree. A path which is not isolated within the > > infinite tree does not exist as isolated path. > > Pray define "isolated". Two path are isolated at level k if they have digits m and n with m =/ = n at a level i <= k. Regards, WM
From: Dik T. Winter on 23 Mar 2007 10:58 In article <1174654681.038185.299380(a)e65g2000hsc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 22 Mrz., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174562837.696384.225...(a)b75g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 21 Mrz., 16:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174481309.873623.92...(a)e1g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > > > > > What portion of the first node > > > > > > is assigned to your path? > > > > > > > > > > How much does the last term of the geometric series contribute to > > > > > the value o the series? This is the value which the first node > > > > > contributes to he path. > > > > > > > > As there is no last term of the geometric series this makes no sense. > > > > > > Nevertheless, it makes sense to calculate the sum of the series. > > > > You can calculate the sum of the series. But what the relation is to the > > contribution of the nodes to the paths is extremely unclear. From your > > statement I derive that the first node contributes nothing to the path. > > the first node contributes exactly as much as the last term of the > geometric series contributes to the sum 2 of the series. As this term > does not exist, it does not contribute anything. The next one before > the last one also does not exist and not contribute. But somehow some > nodes manage to contribute enough to obtain the result 2 after all. How do you obtain that result? You have not shown a proof at all for it. > If we exchange infinitely many terms of the geometric series, its sum > remains 2, because it is absolutely converging. > > As the geometric series contains only the smallest possible infinity > of terms, it should not cause problems to read it from behind. If there were a last one. As there is not a last one I have some difficulty with it, because I do not know where to start. > > > > At every node in the tree uncountably many paths go to the left and > > > > uncountably many paths go to the right. I do now know what you are > > > > meaning here. > > > > > > At no node in the tree there exists a single path. This means: There > > > are no single paths. > > > > Still unclear. Through each node go uncountably many paths. But what you > > mean with "there are no single paths" is unclear. Every two individual > > paths diverge at some node from each other. > > Consider all of them simultaneously. You like to do so when Cantor's > diagonal proof is concerned. Every exchanged digit is followed by > infinitely many digits which have to be exchanged. Nevertheless you > say, it is possible to exchange all of them at one time. Consider the > paths o he tree - all at the same time. Yes, I do. What now? Every two paths separate from each other at some specific node, through all nodes go uncountably any paths, there is no level where all paths do separate. > This is the big mistake of set theory. The unreasonable allowance or > prohibition of instantaneously possible actions. Cantor: yes. Tree: > no. Why? Why not? I do it and come to the same conclusion. > > > > > In the whole tree there can be no more separated paths than nodes. > > > > > > > > Pray give a *proof*, not just a statement. > > > > > > The number of separated paths up to level n is given by the number of > > > nodes of level n. No level of the tree has an uncountable number of > > > nodes. > > > > Makes no sense at all and is no proof. At each level n there are 2^(n-1) > > separated groups of paths, where each group contains uncountably many > > paths. > > That is wrong, because none of these paths ever gets isolated. Is it > similar to the quarks? Path-confinement? How do you define "isolated"? By the common meaning of the word I would say that indeed no two paths are isolated from each other because they have a common node. So you apparently do mean something completely different. > We can safely state that at no level there are uncountably many > separated paths. What do you *mean* with uncountably many separated paths. > As every node is in the tree ad no node is outside, there are at no > place in the world uncountably many numbers. Yes, that is what has been stated all the time. The number of *terminating* paths is countable. What this has to say about non-terminating paths escapes me. > > > > > No. It means that the number of paths which consist only of nodes > > > > > with natural indexes is countable. > > > > > > > > Not at all. Pray provide a *proof*. > > > > > > The number of separated paths up to level n is given by the number of > > > nodes of level n. No level of the tree has an uncountable number of > > > nodes. > > > > Yes, so what? That is no proof of your statement. When we talk about the > > set of paths, we are *not* talking about any finite level. What happens at > > finite levels is irrelevant to the total result. > > Wrong. Every number has digits only at finite distance from the > decimal point. > In the tree there is no level "infinite", but all the infinitely many > levels are there - each one in a finite distance from the root node. Yes, so what? That is still no proof of your statement. What happens at finite levels is irrelevant to the number of non-terminating paths. > > > > > The "distance" is measured by the index of the due node, i.e., by > > > > > the number n of the corresponding level. "At no finite distance > > > > > from the root" means at no node which can be enumerated by a > > > > > natural number. > > > > > > > > Right. There is *no* node where all non-terminating paths are > > > > separated from each other. > > > > > > There is no point in the tree where more than countably many paths are > > > separated, although the tree is infinite. Outside of the tree there is > > > no mathematics of real numbers. > > > > As at each point in the tree there are uncountably many paths going > > through it, I wonder what you mean with "countably many paths are > > separated". > > At no point of the tree (and of the universe) more than countably many > path can be distinguished. I would state that as "at no node in the tree more than countably many groups of paths can be distinguished", or, equivalently "at no node in the tree terminate more than countably many terminating paths". This still does not say anything about the number of non-terminating paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Mar 2007 11:10 In article <1174655745.265882.25720(a)e65g2000hsc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 22 Mrz., 16:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > We can form the union of the sets. Then we get a set of finite paths. > > > > Yes, we can do that. And then we get a set of finite paths that does *not* > > contain a non-terminating path as an element, and also not the path-length > > of such a path as an element. So what now? > > This set of paths forms a tree which does not differ by any node from > the complete tree T(oo). Indeed. As a set of nodes they can not be distinguished. The only difference is that (by your definition), T(oo) contains non-terminating paths while that set of paths does *not* contain non-terminating paths. If we look at nodes as pairs of indices (i, j), where i is the level (from 1 onwards) and j the node in the level (from 1 to 2^(i-1)), the tree is the set: {(i, j_i) | i in N, j_i in {1, ..., 2^(n-1)}} a terminating path is the set (for any natural n): {(i, j_i) | i in {1, ..., n}, j_0 = 1, j_k = 2*j_k-1 or 2*j_k} a non-terminating path is the set: {(i, j_i) | i in N, j_0 = 1, j_k = 2*j_k-1 or 2*j_k} > > Again "contain". But, so what? I fail to see the relevance to the > > path-length of non-terminating paths. > > To spell it out clearly: If you take the infinite union U(T(n)) of all > finite trees T(n), then you do not have any infinite path. Why not? The infinite union is just an infinite set of nodes. A path in a tree is a subset of that infinite sets. There are infinite subsets of that set of nodes that *do* form an infinite path, see above. Still, what is the relevance of path-length? > > > > It is not an element of the union of sets of finite paths. Why do you > > > > think that I do think so? The union of sets of finite paths is a set > > > > of finite paths, so there is no infinite path in it as element. > > > > > > Therefore, after having the union of all paths which are elements of > > > finite trees, there is no infinite path. > > > > Wrong. In the first place, paths are *not* elements of trees according to > > your definitions. Paths are *subsets* of trees according to them. Pray > > use the correct wording. > > OK. Subsets. > > > Second, the union of all paths etc., is *not* a > > path, it is just an infinite set of nodes, the same as the union of all > > trees. The paths are in it as subsets, as are the infinite paths. > > The union of all *finite* paths gives the same infinite set of nodes > as the union of all finite trees. But there is no infinite path > involved / required. Not in the uniting, no. But the union *does* contain infinite paths. Just like the union of all finite initial segments of the natural numbers does have as a subset an infinite initial segment of the natural numbers. > > > > The path-length > > > > is the cardinal number of the set that represents a path (actually > > > > one less if the path is finite). > > > > > > No. You are in error. The cardinal number of an infinite set of finite > > > paths is infinite. But that does not produce any path of infinite > > > length. > > > > But I do not state that it does. The infinite paths are *unions* of finite > > paths, where each finite path is an initial segment of the infinite path. > > I am not talking about the cardinality of a set of path, but of the > > cardinality of a path (the number of nodes in it). That does not mean > > that an infinite set of paths contains an infinite path, that would be > > false indeed. > > The union of all finite paths is an infinite union of finite paths, > but not an infinite path. It is not even a path. It is the infinite tree. > The set of all negative unit fractions is infinite but neither any > member nor the whole set does cover zero. What is the relevance? > > > > At each node, uncountably many paths go to the left and uncountably > > > > many paths go to the right. > > > > > > And everything which belongs to the mathematics of real numbers of [0, > > > 1] happens within the tree. A path which is not isolated within the > > > infinite tree does not exist as isolated path. > > > > Pray define "isolated". > > Two path are isolated at level k if they have digits m and n with m =/ > = n at a level i <= k. So each two non-terminating paths are isolated from each other. And now what? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 23 Mar 2007 11:35 On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174654681.038185.299...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 22 Mrz., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1174562837.696384.225...(a)b75g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 21 Mrz., 16:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1174481309.873623.92...(a)e1g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > > > > > > What portion of the first node > > > > > > > is assigned to your path? > > > > > > > > > > > > How much does the last term of the geometric series contribute to > > > > > > the value o the series? This is the value which the first node > > > > > > contributes to he path. > > > > > > > > > > As there is no last term of the geometric series this makes no sense. > > > > > > > > Nevertheless, it makes sense to calculate the sum of the series. > > > > > > You can calculate the sum of the series. But what the relation is to the > > > contribution of the nodes to the paths is extremely unclear. From your > > > statement I derive that the first node contributes nothing to the path. > > > > the first node contributes exactly as much as the last term of the > > geometric series contributes to the sum 2 of the series. As this term > > does not exist, it does not contribute anything. The next one before > > the last one also does not exist and not contribute. But somehow some > > nodes manage to contribute enough to obtain the result 2 after all. > > How do you obtain that result? You have not shown a proof at all for it. > > > If we exchange infinitely many terms of the geometric series, its sum > > remains 2, because it is absolutely converging. > > > > As the geometric series contains only the smallest possible infinity > > of terms, it should not cause problems to read it from behind. > > If there were a last one. As there is not a last one I have some > difficulty with it, because I do not know where to start. You need not to start. Reverse all terms of the series simultaneously. > > > > > > At every node in the tree uncountably many paths go to the left and > > > > > uncountably many paths go to the right. I do now know what you are > > > > > meaning here. > > > > > > > > At no node in the tree there exists a single path. This means: There > > > > are no single paths. > > > > > > Still unclear. Through each node go uncountably many paths. But what you > > > mean with "there are no single paths" is unclear. Every two individual > > > paths diverge at some node from each other. > > > > Consider all of them simultaneously. You like to do so when Cantor's > > diagonal proof is concerned. Every exchanged digit is followed by > > infinitely many digits which have to be exchanged. Nevertheless you > > say, it is possible to exchange all of them at one time. Consider the > > paths of he tree - all at the same time. > > Yes, I do. What now? Every two paths separate from each other at some > specific node, through all nodes go uncountably any paths, there is no > level where all paths do separate. Because there is no "all paths". > > > > At no point of the tree (and of the universe) more than countably many > > path can be distinguished. > > I would state that as "at no node in the tree more than countably many > groups of paths can be distinguished", That is where we agree. or, equivalently "at no node in > the tree terminate more than countably many terminating paths". No path terminates. This > still does not say anything about the number of non-terminating paths. That is where we disagree. All paths belong to groups of paths. Even single paths belong to groups and are groups, if isolated. I think we have cleared our positions sufficiently: We agree in: At no level in the tree more than countably many groups of paths (including single paths) can be distinguished. And outside of he tree there are no paths. You nevertheless believe in the uncountable while I do not. Regards, WM
From: mueckenh on 23 Mar 2007 11:38
On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Pray define "isolated". > > > > Two path are isolated at level k if they have digits m and n with m =/ > > = n at a level i <= k. > > So each two non-terminating paths are isolated from each other. And now > what? Now we see that their number is countable, because there are never more than countably many paths-bundles including such with only one path isolated from one another. Regards, WM |